Positive Reynolds operators and derivations on C(K)

Andreas Neeb

Abstract

In this paper, it is shown that the spectrum of any positive Reynolds operator is contained in the disc centered at \frac{1}{2} with radius \frac{1}{2}. Also, it is proved that T is injective, provided T is a positive Reynolds operator with dense range. Using this result, a lemma of Miller can be stated as follows: T is a positive Reynolds operator with dense range if and only if D = I −T^{−1} is a closed, densely-defined derivation, which is not bounded, in general. With this relation, a new criterion for a closed, densely-defined derivation D is developed, which is equivalent to D being the generator of a one-parameter semigroup. Namely, the following are equivalent: (1) D is a generator, and (2) 1 ∈ ρ(D) and T = R(1, D) is a positive operator. An alternative version of (2) is given, which is formulated only in terms of the derivation. Statement (2) above yields a representation of positive Reynolds operators T possessing dense range in terms of the semigroup generated by D. As an application, it is obtained that the spectrum of T is either the disc C centered at \frac{1}{2} with radius \frac{1}{2} or σ(T) is a subset of the boundary of C.

0 Introduction

Andreas Neeb

Over the last 40 years, a number of mathematicians published results on Reynolds operators, in particular, name Dubreil-Jacotin [4], Molinaro [12], Rota [14], Billik [2] and Miller [9], [10], [11], [6], [5], who first noted the relationship D = I − T^{−1} between injective Reynolds operators and derivations explicitly. Interestingly, only in the early papers of Dubreil-Jacotin [4] and Molinaro [12], the positivity of operators played a role, whereas their successors did not use it, even though most known examples of Reynolds operators are easily seen to be positive. Much later Scheffold [17] rediscovered this property in the context of Reynolds operators. In this paper, we also consider mostly positive Reynolds operators, since the aspect of positivity yields interesting results which are not true for arbitrary Reynolds operators. In particular, we improve a result of Rota [14] and Billik [2] on the representation of Reynolds operators. In doing so, we reveal the close relation between positive Reynolds operators and derivations which generate a one-parameter semigroup. In addition, for a derivation D, we obtain a necessary and sufficient condition for D to be a generator. Throughout this paper, we consider the Banach spaces C(K) or C_{0}(X) as vector spaces over the real or complex field, which will be specified where necessary. We will always assume that K is a compact and X a locally compact Hausdorff space. Under these assumptions, C(K) is the space of all continuous functions f : K → ℝ (ℂ), and C_{0}(X) is the space of all bounded continuous functions g : X → ℝ (ℂ) vanishing at infinity. Recall that the spaces are both commutative Banach lattice algebras and C^{∗}-algebras, if equipped with supremum norm, pointwise order and multiplication. For brevity, we prove most of our results on C(K), whenever the locally compact case needs special treatment. In fact, most results are true for C_{0}(X) as well, only the proofs require some technical knowledge about locally compact spaces, which cannot yet be found in the literature. Anyhow, the proofs for the locally compact case do not give much more insight, since the ideas are essentially the same as in the compact case. Furthermore, the results are new for C(K), whence we will spare the locally compact case for later publication. First, we define the object of our considerations.

0.1 Definition. A Reynolds operator is a continuous linear operator T on B, a commutative Banach algebra, satisfying the following algebraic condition

(R) T (fg) = T fT g + T ((f − T f)(g − T g)), f, g ∈ B.

The algebraic identity (R) is called Reynolds identity.

In the sequel, we will mostly consider real operators, even if the space C_{0}(X) is assumed to be complex. Recall that a linear operator T is called real, if and only if T maps real elements to real elements, where an element f ∈ C_{0}(X) is called real, if f = \overline{f}, the complex conjugate of f. Note that this means that T is the complexification of an operator on the real space C_{0}(X)_{ℝ}, compare Schaefer, [16, II.11]. In particular, positive operators are real. Of course, this means no restriction, if T is given on the real space C_{0}(X)_{ℝ}. Furthermore, we note that the complexification of a Reynolds operator on C_{0}(X)_{ℝ} yields a real Reynolds operator on the complex space C_{0}(X), which can be quickly verified. As a first well-known result on Reynolds operators, we state the fact that the range of a Reynolds operator is a subalgebra.

0.2 Lemma. Let T be a continuous Reynolds operator on C_{0}(X). Then the range A of T is a subalgebra of C_{0}(X). If T is real, then A is invariant under conjugation. For the closure \overline{A} of A we have that kerT · \overline{A} ⊆ kerT. Consequently, if A is dense in C_{0}(X), then the kernel of T is an algebraic (and lattice) ideal of C_{0}(X).

Proof. This statement follows directly from the Reynolds identity and a completeness argument.

The following result is due to Scheffold [17].

0.3 Proposition. Every positive Reynolds operator T on C_{0}(X) has norm less than or equal to 1.

Rota [14] proved the subsequent result, which will be used frequently in the sequel.

0.4 Proposition. Let T be a real Reynolds operator on C_{0}(X).

(1) For any n ∈ ℕ and every f ∈ C_{0}(X) we have

nT (f(T f)^{n−1}) = (n − 1)T ((T f)^{n}) + (T f)^{n}.

(2) Let q : ℝ → ℝ be a continuously differentiable function satisfying q(0) = 0. For any real f ∈ C_{0}(X), we have q ∘ f = q(f) ∈ C_{0}(X) and

T (q(T f)) = q(T f) + T (T fq′(T f)) − T (fq′(T f)).

Furthermore, for any real f in the range of T, we have that q(f) is contained in the range of T as well.

Positive Reynolds operators and derivations

Another important tool is the following relation between Reynolds operators and derivations, which was noted by Miller [9]. Recall that a derivation D on an algebra A is an operator satisfying D(ab) = aDb + (Da)b, for any a, b ∈ A. We call a derivation D a ∗-derivation if it is real, that is, D\overline{f} = \overline{Df} for f ∈ A. This terminology follows Sakai [15], where many important results on derivations can be found. Note that the derivations we are concerned with are unbounded, in general.

0.5 Proposition. Let T be a continuous, injective Reynolds operator on C_{0}(X) with dense range. Then D = I − T^{−1} is a densely defined and closed derivation with 1 ∈ ρ(D), the resolvent set. Let on the other hand D be a closed, densely defined derivation such that λ ∈ ρ(D). Then T = λR(λ, D) = λ(λI − D)^{−1} is a continuous, injective Reynolds operator with dense range. Furthermore, if λ = 1 ∈ ρ(D), the following dual relation between D and T = R(1, D) is valid: D = I − R(1, D)^{−1} and T = R(1, I − T^{−1}).

Let T be a Reynolds operator with the properties according to Proposition 0.5. Because of the relation, we will call D = I − T^{−1} the derivation associated to T. Conversely, for a closed, densely defined derivation D with 1 ∈ ρ(D), we will call T = R(1, D) the Reynolds operator associated to D. Note that in this situation, T is real, if and only if its associated derivation D is a ∗-derivation. Clearly, if D is a derivation on C(K) with e contained in the domain of D, then De = 0, where e denotes the unit of C(K). Then, if D satisfies the prerequisites of Proposition 0.5, the associated Reynolds operator T is Markovian, that is, T e = e. For derivations, the following chain rule is valid, just as in the case of the first derivative. This well-known result can be found in Sakai [15], Proposition 3.2.3. Note that this result is equivalent to statement (2) of Proposition 0.4, if D is densely defined and 1 ∈ ρ(D).

0.6 Lemma. Let D : A → C_{0}(X) be a closed ∗-derivation and let g ∈ A be a real element. If q : [−∥g∥, ∥g∥] → ℝ is a differentiable continuous function with q(0) = 0, then, the composite q ∘ g lies in A and D satisfies the chain rule, that is,

D(q ∘ g) = (q′ ∘ g) Dg.

1 On the injectivity of positive Reynolds operators

In this section, we prove that a positive Reynolds operator T is injective on its range. For positive Reynolds operators with dense range, we improve this result and show that T is injective. In the following, we use the pointwise Reynolds identity introduced by Scheffold [17]. This identity is given by

(R^{∗}) ∫_{K}(f − T f)T g + gT f − (T f)(t)g dμ_{t} = 0 for t ∈ K,

which is equivalent to the Reynolds identity if we set μ_{s} = T′δ_{s}, for the adjoint operator T′ and the point evaluation δ_{s} in the dual of C(K). This identity proves to be very useful in the sequel.

In the following, we prove the results only for the real Banach lattice C(K)_{ℝ}. For the complex case we use the complexification theory of vector lattices, which can be found in Schaefer [16, II.11]. This theory gives that there is a one-to-one correspondence between positive operators on E_{ℝ} and E_{ℂ}. Let T be a positive operator on E_{ℝ}, then its complexification T_{ℂ} : E_{ℂ} → E_{ℂ} is given by the canonical extension T_{ℂ}(u + iv) = T u + iT v (u, v ∈ E_{ℝ}). It is easy to see that properties like positivity, injectivity or density of the range are invariant under the complexification process, which implies that the following results are valid in the complex case as well.

1.1 Proposition. Let T be any continuous Reynolds operator on C(K)_{ℝ} and let A denote the range of T. If T is positive then T restricted to A is injective.

Proof. For g ∈ A, there is an h ∈ C(K)_{ℝ} satisfying T h = g. We have to show that T^{2}h = 0 implies T h = 0. We set f := T h and g := h, then the pointwise Reynolds identity yields

∫_{K}(T h)^{2} dμ_{t} = 0 for t ∈ K.

The positivity of T implies that T′ is positive. Hence the measure μ_{t} = T′δ_{t} is also positive, and the equation above implies T h ≡ 0 on the support supp(μ_{t}) for any t ∈ X. Now let t ∈ K be fixed. It remains to show that T h ≡ 0 on supp(μ_{t}) implies T h(t) = 0. This is achieved by setting f := g := h in (R^{∗}), as follows:

0 = ∫_{K}(2hT h − (T h)^{2} − (T h)(t)h) dμ_{t}
= −(T h)(t) ∫_{K} h dμ_{t}
= −[(T h)(t)]^{2},

as ∫_{K} h dμ_{t} = T h(t). Thus we have accomplished T h(t) = 0 for any given t ∈ K. But this means that T h ≡ 0 which is what we had to show.

1.2 Theorem. Let T be a positive Reynolds operator on C(K)_{ℝ} with dense range A. Then T is injective.

Proof. According to Lemma 0.2, the kernel of T is a closed lattice and algebraic ideal. By a result of Stone [18], there is a closed subset L ⊆ X such that for any element f ∈ C(K)_{ℝ} the following statements are equivalent: (1) f ∈ ker T and (2) f(t) = 0 for all t ∈ L. Let now L^{c} = K \ L and let M be any closed (hence compact) subset of L^{c}. Then M and L are two disjoint closed subsets of K. In order to obtain a contradiction we assume that M ≠ ∅. Therefore, it is possible to find an element f ∈ range T such that f|_{M} ≥ 1 and f|_{L} ≤ 0. Let us consider a continuously differentiable function q : ℝ → ℝ, 0 ≤ q ≤ 1 which satisfies q(t) = 0 for t ≤ 0 and q(t) = 1 for all t ≥ 1. According to Proposition 0.4, we have that q ∘f is contained in the range of T. On the other hand, q ∘f(t) = 1 for t ∈ M and q ∘f(t) = 0 for t ∈ L. Thus q ∘f ∈ kerT ∩ rangeT. According to Proposition 1.1, T is injective on its range, thus kerT ∩ rangeT = {0}. Hence q ∘ f = 0 which gives a contradiction to M ≠ ∅. Thus M must be empty, hence L = K, which is equivalent to the injectivity of T.

2 Spectral properties of Reynolds operators and derivations

In this section, we locate the spectrum of a Reynolds operator and its associated derivation. Furthermore, we give a formula which relates the resolvents of T and D. If the given space C_{0}(X) is real, we have to say what we mean by the spectrum of an operator. In order to do this, we consider the canonical complexifications of C_{0}(X) and T, as mentioned at the beginning of the previous section. We will not distinguish between T and its complexification by change of notation, by the convention that we always mean the spectrum of T_{ℂ}, if C_{0}(X) is real.

2.1 Theorem. Let T be a positive Reynolds operator on C_{0}(X). Then the spectrum σ(T) is contained in the disc C = {z ∈ ℂ: |z − 1/2| ≤ 1/2}.

Proof. First, we consider the canonical biadjoint T′′ : C(K) → C(K), where C(K) is the bidual of C_{0}(X). Recall that the spectra of T and T′′ coincide, since an operator S : B → B (B Banach space) is continuously invertible if and only if S′′ : B′′ → B′′ is invertible. We consider the F -product Ẽ of countably many copies of the spaces C(K) with respect to a filter F finer than the Frechet filter (that is, all sets M ⊂ ℕ with finite complements, compare Schaefer [16, V.1]). Let T̃ be the canonically defined extension of T′′ on Ẽ. This operator also satisfies the Reynolds identity. Again, the spectra of T′′ and T̃ coincide, according to the construction of the F -product. By this procedure, all the approximate eigenvalues of T′′ are transformed into proper eigenvalues of T̃. In particular every value in the topological boundary of the spectrum of T̃ is an eigenvalue. Recall that the F -product is again a space C(Ω), where Ω is compact. Now let λ ∈ σ(T̃) be any eigenvalue to the eigenvector f ≠ 0, that is, T̃f = λf. From the positivity of T̃ we get T̃f̄ = λ̄f̄, taking complex conjugates. Without loss of generality we may assume that λ ≠ 0. Using the Reynolds identity and the fact that ff̄ ≠ 0 and positive, we obtain

(λ + λ̄ − λλ̄)T̃(ff̄) = λλ̄ ff̄.

The scalars in the equation are real, which allows to give the estimate

0 ≤ \frac{λλ̄}{λ + λ̄ − λλ̄} ≤ 1,

according to Proposition 0.3. Taking λ = α + iβ leads to

0 ≤ \frac{α^{2} + β^{2}}{2α − (α^{2} + β^{2})} ≤ 1.

This gives 2α > α^{2} + β^{2} for the denominator and α^{2} + β^{2} ≤ α, as the numerator is less than or equal to the denominator. Clearly, the second inequality implies the first one. The second inequality is equivalent to

(α − \frac{1}{2})^{2} + β^{2} ≤ \frac{1}{4}.

This inequality determines exactly the closed disc centered at \frac{1}{2} with radius \frac{1}{2}, that is, C. Hence the boundary of the spectrum of T′′ is contained in C, which implies the assertion.

In Example 4.2, we exhibit a Reynolds operator T with σ(T) = C which implies that the given estimate is sharp. For a positive Reynolds operator with dense range, we can locate the spectrum of D, the associated derivation. This is achieved using the Spectral Mapping Theorem for unbounded operators by Greiner (see Nagel [13, Prop. A.III.2.5]).

2.2 Lemma. Let T : C(K) → C(K) be a positive Reynolds operator with dense range, and let D = I − T^{−1} be the associated derivation. Then the spectrum of D is contained in the left half-plane {λ ∈ ℂ: Reλ ≤ 0}. Proof. Note that T is injective, according to Theorem 1.2, hence D is a well-defined and densely defined operator. The closedness of D follows from the injectivity of T. By the prerequisite, R(1, D) = T is a bounded operator, which implies that 1 is in the resolvent set of D. Now we can use Greiner’s Proposition [13, Prop. A.III.2.5]. This result states that for a closed, densely defined operator D on a Banach space, given λ_{0} ∈ ρ(D) and α ∈ ℂ \ {λ_{0}}, the point α is contained in σ(D) if and only if (λ_{0} − α)^{−1} ∈ σ(R(λ_{0}, D)). We use this result, taking λ_{0} = 1 and α ∈ \{z ∈ ℂ: |z − \frac{1}{2}| ≤ \frac{1}{2}\}, according to the location of the spectrum of T (Theorem 2.1). Thus we conclude that (1 − α)^{−1} ∈ σ(T) if and only if α ∈ σ(D). To complete the proof, it is only necessary to check that the inverse function β ↦ 1 − β^{−1} maps the set \{z ∈ ℂ: |z − \frac{1}{2}| ≤ \frac{1}{2}\} \ \{0\} onto \{λ ∈ ℂ: Reλ ≤ 0\}.

In the sequel, we establish a relation between the resolvent of D and an operator which is related to the resolvent of T. This is used to prove that λR(λ, D) is positive for 0 < λ < 1.

2.3 Proposition. Let T be a positive Reynolds operator on C(K) with dense range A and let D = I − T^{−1} be the derivation associated to T. Then the following statements are valid:

(1) For any λ ∈ {z ∈ ℂ: Re(z) > 0 and z ≠ 1} we have

\frac{1}{1−λ}R(\frac{1}{1−λ}, T)T = R(λ, D).

(2) For 0 < λ ≤ 1 the resolvent R(λ, D) is positive and, in particular, ∥λR(λ, D)∥ ≤ 1. Proof. (1) Note that T is injective, according to Theorem 1.2. The resolvent R(λ, D) exists by Lemma 2.2. For the same reason, \frac{1}{1−λ} is not in the spectrum of T, as |(1 − λ)^{−1} − 1/2| > 1/2. Thus the statement follows from

\frac{1}{1−λ}(\frac{1}{1−λ}I − T)^{−1}T = \frac{1}{1−λ}(\frac{1}{1−λ}T^{−1} − T^{−1}T)^{−1}
= (\frac{1}{1−λ}(1 − λ)T^{−1} − (1 − λ)I)^{−1}
= (T^{−1} − (1 − λ)I)^{−1}
= (λI − (I − T^{−1}))^{−1} = R(λ, D).

(2) Let 0 < λ < 1. For λ ≠ 1 we have |1 − λ|^{−1} > 1 ≥ ∥T ∥, according to Proposition 0.3. Thus, the following identity holds

\frac{λ}{1 − λ}(\frac{1}{1 − λ}I − T)^{−1}T = \frac{λ}{1 − λ} ∑_{k=0}^{∞}(1 − λ)^{k+1}T^{k+1},

from which the positivity of the given operator follows. Therefore, statement (1) implies the positivity of λR(λ, D). According to Proposition 0.5, λR(λ, D) is a Reynolds operator. Hence the norm estimate follows from the positivity and Proposition 0.3. For λ = 1, the result follows from R(1, D) = T.

3 Generating derivations and representation of Reynolds operators

Derivations play a great role as generators of strongly continuous one-parameter semigroups of algebra homomorphisms. The reader is assumed to be familiar with the basics of this important part of mathematics, a thorough treatment can be found in the monographs by Goldstein [7], Hille-Phillips [8] or Nagel [13], the latter contains material on positive semigroups especially. In particular, we use the well-known Theorem of Hille and Yosida (compare Nagel [13, Cor. A.II.1.8]). It states that a densely defined, closed operator H is the generator of a semigroup of contractions if and only if, for all λ > 0, the estimate ∥λR(λ, H)∥ ≤ 1 is valid (in particular ]0, ∞[ ∈ ρ(H)). In the sequel, we prove that this condition is satisfied for derivations if a much simpler condition is fulfilled. This condition is closely related to the positivity of the associated Reynolds operator. In this section, we will again always consider the complex spaces C(K), if not stated otherwise. If the given space is real, we simply consider the canonical complexification, as mentioned at the beginning of Section 1.

3.1 Definition. Let H be a closed, densely defined and real (possibly unbounded) operator on C(K). Then we say that H satisfies the principle of inverse monotony (IM) if and only if

(IM) for any real g ∈ A, Hg ≤ g implies g ≥ 0.

Provided that D is a ∗-derivation with 1 ∈ ρ(D), then property (IM) is equivalent to the positivity of the associated Reynolds operator T = R(1, D).

3.2 Proposition. Let T be a real injective Reynolds operator on C_{0}(X) with dense range A and D = I−T^{−1} the associated ∗-derivation. Then the following statements are equivalent:

(P) T is a positive operator.

(IM) For any real g ∈ A, Dg ≤ g implies g ≥ 0.

Proof. Suppose T is positive and Dg ≤ g holds for a given g ∈ A, g = \overline{g}. Consider f = T^{−1}g. Then we conclude Dg = (I − T^{−1})g = g − f ≤ g which is equivalent to f ≥ 0, and hence T f = g ≥ 0 by the positivity of T. Thus (IM) holds. Conversely, let D satisfy (IM) and let f ≥ 0. Consider g = T f. Thus T^{−1}g = f ≥ 0, and hence Dg = (I − T^{−1})g ≤ g. Now, (IM) gives g ≥ 0, and thus T is positive as required.

We proceed in proving that property (IM) for the ∗-derivation D implies that λD possesses property (IM) for any λ > 0. In the proof below, we use the fact that D is closed and densely defined on A ⊂ C(K), and in particular, that the domain of D contains the constants, which has been noted by Sakai [15, Prop. 3.3.4].

3.3 Proposition. Let D : A → C(K) be a densely defined, closed ∗-derivation with ρ(D) ⊃ {λ ∈ ℝ: λ > 0} and assume that D satisfies (IM). Then μD satisfies (IM) for any μ > 0.

Proof. Clearly, the Reynolds operator T = R(1, D) exists and is positive, by Proposition 3.2. Therefore, D is a ∗-derivation and μD satisfies (IM) for μ > 1, since the Reynolds operators μ^{−1}R(μ^{−1}, D) = R(1, μD) are positive for μ^{−1} ∈ ]0, 1[, according to Proposition 2.3. Therefore, it is only necessary to prove the assertion for μ ∈ ]0, 1[.

First, we assume there exists a real g ≤ 0 in A which satisfies μDg ≤ g. Since 0 < μ < 1, this yields Dg ≤ μDg ≤ g, which in turn implies g ≥ 0. But this gives g = 0, by the assumption g ≤ 0. Hence there cannot exist a non-zero negative function g with μDg ≤ g. Therefore, we assume that there is a non-positive real function g = g^{+} − g^{−} ∈ A with the property that μDg ≤ g. If we can prove the existence of a negative element g_{1} ∈ A, g_{1} ≠ 0, such that μDg_{1} ≤ g_{1}, then we obtain a contradiction since such a g_{1} cannot exist, according to the above argument. The following construction shows that under the assumption, there exists a function g_{1} ≠ 0 with −g^{−} ≤ g_{1} ≤ 0 satisfying μDg_{1} ≤ Dg_{1}. To this end, let v = −1/2 inf{g(x) : x ∈ X}. Then v > 0, according to the assumption g^{−} ≠ 0. We set

q_{v}(r) = \{ \begin{matrix}0 & for r ≥ 0 \\ −v^{−1}r^{2}(v^{−1}r + 2) & for −v ≤ r ≤ 0 \\ r & for r ≤ −v. \end{matrix}

Note that q_{v} is continuously differentiable and

q′_{v}(r) = \{ \begin{matrix}0 & for r ≥ 0 \\ −v^{−1}r(3v^{−1}r + 4) & for −v ≤ r ≤ 0 \\ 1 & for r ≤ −v. \end{matrix}

We proceed setting g_{1}(x) = q_{v}(g(x) + v), explicitly

g_{1}(x) = \{ \begin{matrix}0 & for x with g(x) ≥ −v \\ −v^{−1}(g(x) + v)^{2}(v^{−1}(g(x) + v) + 2) & for x with −2v ≤ g(x) ≤ −v \\ g(x) + v & for x with g(x) ≤ −2v. \end{matrix}

We have to prove that g_{1} ∈ A, the domain of μD (which is of course the domain of D). But this is clear from Lemma 0.6 and from the fact that the domain of any closed derivation contains e, the unit, according to Sakai [15, Prop. 3.3.4]. Furthermore,

Dg_{1} = q′_{v}(g + v e)D(g + ve) = q′_{v}(g + v e)Dg,  (1)

since D is a derivation, and since De = 0. Clearly, g_{1} is negative and not equal to zero. It remains to prove that μDg_{1} ≤ g_{1}. Firstly, let x ∈ K such that g(x) ≥ −v. Then we have g_{1}(x) = 0 and, according to above, Dg_{1}(x) = q′_{v}(g(x) + v) Dg(x). Since q′_{v}(g(x) + v) = 0, according to the choice of x, we have Dg_{1}(x) ≤ 0 ≤ g_{1}(x). Thus, μDg_{1}(x) ≤ g_{1}(x), by the positivity of μ. Secondly, let x ∈ K such that g(x) ≤ −2v. Then we have

g_{1}(x) = q_{v}(g(x) + v) = g(x) + v,

according to the definition of q_{v}. Furthermore, since De = 0, we obtain

Dg_{1}(x) = q′_{v}(g(x) + v) Dg(x) = Dg(x).

Consequently, μDg_{1}(x) = μDg(x) ≤ g(x), according to our assumption. Therefore, we obtain μDg_{1}(x) ≤ g(x) + v = g_{1}(x) for the points x ∈ {y ∈ K : g(y) ≤ −2v} considered in this case.

Thirdly, consider x ∈ K such that −2v < g(x) < −v. Then equation (1) together with relation μDg ≤ g gives

μDg_{1}(x) = μq′_{v}(g(x) + v)Dg(x) ≤ g(x) q′_{v}(g(x) + v).

It remains to prove the following inequality

g(x) q′_{v}(g(x) + v) ≤ q_{v}(g(x) + v) = g_{1}(x).

Explicitly, this is equivalent to showing that

g(x) (−\frac{3}{v^{2}} (g(x) + v)^{2} − \frac{4}{v}(g(x) + v)) ≤ −\frac{1}{v^{2}} (g(x) + v)^{3} − \frac{2}{v}(g(x) + v)^{2}.

Substituting z = g(x) and simplifying yields

(z + v)z(−\frac{3}{v^{2}} (z + v) − \frac{4}{v}) + (z + v)^{2}(\frac{1}{v^{2}} (z + v) + \frac{2}{v}) ≤ 0,

which, since −v < 0 and (z + v)^{−1} < 0, is equivalent to

z(\frac{3}{v}(z + v) + 4) − (z + v)(\frac{1}{v}(z + v) + 2) ≤ 0.

Thus we have a quadratic polynomial p in z which is given by

p(z) = \frac{3}{v}z^{2} + 4z + 3z − (\frac{1}{v}z^{2} + 2z + v + 2z + 2v) = \frac{1}{v}(2z^{2} + 3vz − 3v^{2}).

We claim that p(z) ≤ 0 for z ∈ ] − 2v, −v[. In order to prove this, we determine the roots of p and state that p is a parabola which is negative between the roots and positive outside, since 2/v is positive. The roots are given by

z_{1,2} = −\frac{3}{4}v ± \sqrt{\frac{9}{16}v^{2} + \frac{24}{16}v^{2}} = \frac{1}{4}v( − 3 ± \sqrt{33}).

Now, z_{1} = v/4(−3 + \sqrt{33}) ≥ v/4(−3 + 5) = v/2 and z_{2} = v/4(−3 − \sqrt{33}) ≤ v/4(−3 − 5) = −2v. Hence p is negative in the interval ] − 2v, v/2[, which proves the claim. Therefore, we have proved that μDg_{1}(x) ≤ g_{1}(x), for every x ∈ K with −2v < g(x) < −v. Altogether, we proved μDg_{1} ≤ g_{1}, globally on K, where g_{1} ≤ 0. According to the remark at the beginning of the proof this gives that g_{1} = 0, hence the function g must have been positive in contrast to our assumption. Consequently, μDg ≤ g implies that g is positive, which is equivalent to property (IM) for μD.

Now we are in the situation to prove the main theorem on the generating properties of derivations, since the essential step is completed by Proposition 3.3 above. We first quote a well-known result which is used in the context.

3.4 Remark. Recall that a generator of a one-parameter semigroup is a derivation if and only if the semigroup consists of algebra homomorphisms. This statement can be found in Nagel [13, Thm. B.II.3.4]. Furthermore, every algebra homomorphism is a contraction and a Markovian lattice homomorphism, hence positive. In particular, if C(K) is complex, this implies that D is a ∗-derivation, since D maps real elements to real elements.

3.5 Theorem. (Generation Theorem) Let D be a closed, densely defined derivation on C(K). Then the following statements are equivalent:

(1) D is a ∗-derivation with 1 ∈ ρ(D) satisfying condition (IM).

(2) R(1, D) exists and is positive.

(3) For all λ > 0, R(λ, D) exists and is positive.

(4) D generates a one-parameter semigroup of algebra homomorphisms on C(K).

Proof. (1) ⇒ (2) This implication is immediate from Proposition 0.5 and Proposition 3.2.

(2) ⇒ (1) If R(1, D) is positive, then D is a ∗-derivation, since D maps real elements to real elements. Thus, the statement follows from Proposition 3.2.

(2) ⇒ (3) The prerequisite together with Proposition 0.5 implies that T is a positive Reynolds operator. According to Lemma 2.2, σ(D) is contained in the left half-plane, which implies the existence of the resolvents R(λ, D). We recall that

T_{λ} = λR(λ, D) = R(1, λ^{−1}D)

is an injective Reynolds operator with dense range, according to Proposition 0.5, since λ^{−1}D is a derivation. Statement (2) together with Proposition 3.3 imply that λ^{−1}D satisfies (IM) for all λ > 0. Thus, the positivity of T_{λ} = R(1, λ^{−1}D) follows, according to Proposition 3.2, proving (3).

(3) ⇒ (4) First, we note that T_{λ} = λR(λ, D) is a Reynolds operator, since λ^{−1}D is a derivation. Furthermore, T_{λ} is positive, according to the prerequisite. In order to apply the Theorem of Hille and Yosida, we have to prove that ∥T_{λ}∥ ≤ 1, whenever λ > 0. This follows from Proposition 0.3. Thus, D is the generator of a one-parameter semigroup. The assertion follows from Remark 3.4 above.

(4) ⇒ (2) Let the derivation D be a generator of a one-parameter semigroup τ of algebra homomorphisms. Remark 3.4 implies that τ is positive. The Theorem of Hille and Yosida implies the existence of the resolvents R(λ, D) for λ > 0, since τ is a contraction semigroup. By the positivity of τ and the following formula for the resolvent of a generator (compare Nagel [13, B.II.1(1.1)])

R(λ, D)f = ∫_{0}^{∞} e^{−λs}τ(s)f ds,  (2)

we obtain that T = R(1, D) is positive, which gives the conclusion.

This result implies a few other consequences for derivations. In the literature, some criteria are given, which are necessary for a densely defined closed operator to generate a positive one-parameter semigroup. These conditions are all satisfied for D, if any of the conditions of the Generation Theorem is satisfied. In particular, D is dissipative (see Goldstein [7]) and D satisfies the positive minimum principle (see Nagel [13, B.II.1.5]). In particular, Batty, Bratteli and Robinson [1] proved a result which comes nearest to our Generation Theorem 3.5. They proved the equivalence of (3) and (4) in the more general setting of C^{∗}-algebras using different methods. We emphasize that conditions (1) and (2) seem to be new. In fact, (1) is a condition directly formulated for the derivation. This shows, in particular, that it is not necessary to calculate the resolvents of D in order to decide whether D is a generator or not. On the other hand, condition (2) is a priori a much weaker statement than (3), since the positivity of one bounded operator is much easier to check than that of uncountably many. In the proof of the Representation Theorem 3.6 below, we will see the profitableness of (2). Now, we turn back to positive Reynolds operators with dense range. Together with the previous results on derivations, we are able prove that T possesses an integral representation in terms of the semigroup generated by the associated ∗-derivation D = I − T^{−1}.

3.6 Theorem. (Representation Theorem) Let T be a positive Reynolds operator on C(K) possessing dense range and let D = I −T^{−1} be the associated derivation. Then D is the infinitesimal generator of a semigroup τ of algebra homomorphisms and T can be represented by

T f = ∫_{0}^{∞} e^{−t}τ(t)f dt.  (3)

On the other hand, let τ be a one-parameter semigroup of algebra homomorphisms on C(K) with generator D. Then D is a derivation and T = R(1, D) exists. Furthermore, T is a positive Reynolds operator with dense range which can be represented by the semigroup via formula (3) above.

Proof. Let T : C(K) → C(K) be a positive Reynolds operator with dense range. Then the associated derivation D is closed and densely defined, according to Proposition 0.5. Since T = R(1, D) is positive, condition (2) of the Generation Theorem 3.5 implies that D is the generator of a one-parameter semigroup τ of algebra homomorphisms. Using formula (2) from the proof of Theorem 3.5 for λ = 1, we obtain

T = R(1, D) = ∫_{0}^{∞} e^{−t}τ(t) dt,

which is the desired representation of T. Conversely, let τ be given. Clearly, the generator D of τ is a closed, densely defined derivation, according to Remark 3.4. Thus, T = R(1, D) is a positive operator, by the Generation Theorem 3.5. From Proposition 0.5 we obtain that T is a Reynolds operator with dense range. Finally, the first part of the proof gives the representation of T.

At this moment, it seems appropriate to point out that the new condition (2) of Theorem 3.5 enables us to prove the Representation Theorem in the given form, without presupposing that D = I − T^{−1} is the generator of a semigroup. In this context, we note that a variant of the result of the Representation Theorem has been proved by Rota [14] in the setting of spaces L^{∞}(X, Σ, μ) with finite measure. Later, Billik [2] proved that T : C(K) → C(K) possesses a representation as given above if and only if T satisfies a technical condition which is equivalent to the dissipativity of D = I − T^{−1}. He did not notice, however, that positivity together with density of the range is sufficient for the representation. Finally, we note that we do not know whether it is possible to weaken the hypothesis concerning the density of the range, since it is not known whether it is true for a positive Markovian Reynolds operator, that the restriction T |\overline{A} (A = rangeT) has dense range in \overline{A}. If this was true, then we were able to generalize the Representation Theorem 3.6, yielding a representation of T on \overline{A}, the closure of the range of T, since \overline{A} is isomorphic to a space C(L), L compact. Another consequence of an affirmative answer to this question was, that the restriction T |\overline{A} would always be injective, improving Theorem 1.2. From the theory of one-parameter semigroups, we obtain more information about the semigroup which represents a positive Reynolds operator with dense range.

3.7 Remark. In the Representation Theorem 3.6, the semigroup consists of Markovian lattice homomorphisms, as these are exactly the same as algebra homomorphisms, which is a well-known fact (compare Schaefer [16]). Furthermore, Theorem B.II.3.4 of Nagel [13] states that the semigroup τ is given by a semiflow on K, that is, a function ϕ : [0, ∞[×K → K satisfying φ(s + t, x) = φ(s, φ(t, x)). The relation between the semigroup and the semiflow is given by

τ(t)f(x) = f(φ(t, x)).

This gives a powerful tool for the determination of the semigroup if a representable Reynolds operator is given. This will be used in Example 4.2 below.

4 Applications and examples

From the Representation Theorem 3.6, we obtain the following characterization of the spectrum of a positive Reynolds operator on C(K) with dense range. In order to obtain the desired result, we use a statement of Derndinger and Nagel [3] on the spectrum of a generating derivation.

4.1 Theorem. Let T be a positive Reynolds operator on C(K) with dense range. Then either the spectrum of T coincides with the disc

C = \{λ ∈ ℂ: |λ − \frac{1}{2}| ≤ \frac{1}{2}\},

or is contained in the set

Γ = \{λ ∈ ℂ: |λ − \frac{1}{2}| = \frac{1}{2}\}.

Proof. The Reynolds operator T possesses a representation in terms of a one-parameter semigroup of algebra homomorphisms, according to Theorem 3.6. In particular, the associated derivation D is a generator. Thus we are in a situation where we can use Theorem 4.2 of Derndinger and Nagel [3], which states, that there are two alternative cases for the spectrum of a generating derivation D on C(K), namely (1) σ(D) = {λ ∈ ℂ: Reλ ≤ 0} or (2) σ(D) is a union of subgroups of the imaginary axis. We use the Spectral Mapping Theorem of Greiner (compare Nagel [13, Prop. A.III.2.5]) for unbounded operators to obtain the conclusion. The image of D under the mapping

f(z) = \frac{1}{1 − z}

is T, thus the spectrum of D is mapped onto σ(T) \ {0}. In the first case, the left half plane is mapped onto C \ {0}, from which the statement follows, since σ(T) is compact. For the second case, an argument of the same type gives the desired conclusion.

This gives a simple criterion to determine which of the given alternatives is valid for the spectrum of T, a positive Reynolds operator with dense range. It suffices to check whether the point λ = 1/2 is in the spectrum of T in order to decide between the two alternatives. Finally, let us consider an example of a Reynolds operator, which illustrates the applications of the theory developed above.

4.2 Example. Let T on E = C([0, 1]) be given by

T f(s) = ∫_{0}^{1} f(ts) dt for s ∈ [0, 1].

Obviously, T is a positive convolution operator on the unit interval. According to Scheffold [17], T is also a Reynolds operator. Note that, the spectrum of T is the whole disc C = {λ ∈ ℂ : |λ − 1/2| ≤ 1/2}. This can be seen as p_{1}(x) = x is an eigenvector to λ = 1/2, using Theorem 4.1. In fact, it is not difficult to see that any monomial p(x) = x^{r} is an eigenvector to the eigenvalue 1/(1 + r), as long as Rer > 0. Clearly, T has dense range, as T e = e and since the polynomials separate the points of [0, 1], according to Lemma 0.2 and the Theorem of Stone and Weierstraß. Thus T possesses dense range, hence is injective, according to Theorem 1.2. By substitution, we obtain the following representation for T f(s), where s ≠ 0

T f(s) = \frac{1}{s} ∫_{0}^{s} f(t) dt,

and T f(0) = f(0). The range of T is dense, according to the above, and contained in the set of all f ∈ C([0, 1]) which are continuously differentiable on ]0, 1], that is, A ⊆ C([0, 1]) ∩ C^{1}(]0, 1]). Thus, the operator T^{−1} : A → C(K) is well-defined. This operator is given by

T^{−1}g(s) = g(s) + sg′(s).

By Proposition 0.5, the associated derivation is given by D = I − T^{−1}, which yields

Dg(s) = −sg′(s) (s ∈ [0, 1]).

According to the positivity of T, the derivation D satisfies (IM). Hence D is the generator of a one-parameter semigroup τ of algebra homomorphisms, according to the Generation Theorem 3.5. In particular, the operators λR(λ, D) are positive for all λ > 0. In this example, we can give an explicit formula for these Reynolds operators, by,

λR(λ, D)f(s) = \frac{λ}{s^{λ}} ∫_{0}^{s} f(t)t^{λ−1} dt.

This identity is easily verified by applying the operator (λI − D) to the integral on the right hand side. Now, the natural question is the one for a representation of the operators τ(t). According to Remark 3.7, a semigroup of algebra homomorphisms admits a representation by τ(t)f(x) = f(φ(t, x)) where φ : ℝ_{+} × K → K is a semi-flow on the set K (in our case [0, 1]). Our aim is to determine this semi-flow. According to the theory of one-parameter semigroups, we obtain

−sf′(s) = Df(s) = lim_{t→0} \frac{f(φ(t, s)) − f(s)}{t} = \frac{d}{dt}f(φ(t, s))|_{t=0}
= [f′(φ(t, s)) · \frac{d}{dt}φ(t, s)]_{t=0}
= f′(s) · [\frac{d}{dt}φ(t, s)]_{t=0} ,

since φ(0, s) = s. Thus \frac{d}{dt}φ(t, s)|_{t=0} = −s. We conjecture that φ(t, s) = se^{−t}. This is proved by taking the semigroup τ′ given by τ′(t)(f) = f(se^{−t}) and showing that its generator is D, the generator of τ.

\frac{d}{dt}τ′(t)(f) = lim_{t→0} \frac{1}{t}(τ′(t) − I)(f)
= lim_{t→0} \frac{1}{t}(f(se^{−t}) − f(s))
= f′(se^{0}) · (−1)se^{0} = −sf′(s) = Df(s).

Thus the generator of the semigroup τ′ is D, as required. According to the one-to-one correspondence between one-parameter semigroups of contractions and their generators, the semigroups τ′ and τ must be identical. Consequently, the semigroup τ is given by

τ(t)f(s) = f(s · e^{−t}) t ∈ ℝ_{+}.

Finally, we are able to give a representation of the given operator T in terms of the associated semigroup τ. This can be done, applying the Representation Theorem 3.6. Thus the desired representation is

T f(s) = ∫_{0}^{∞} e^{−t}[τ(t)f](s) dt = ∫_{0}^{∞} e^{−t}f(se^{−t}) dt.

Acknowledgment. The author wishes to express his sincere thanks to Prof. Scheffold for many valuable hints and discussions, and, not the least, for his encouragement.

References

[1] Batty, C.J.K., Bratteli, O. & Robinson, D.W.: The heat semigroup, derivations, and Reynolds’s identity, in: “Operator algebras and applications” Vol. 2, Mathematical physics and subfactors, UK-US joint seminar Warwick, UK 1987, London Math. Soc. Lecture Notes Ser. 136 (1988), 23–47.

[2] Billik, M.: Idempotent Reynolds operators, J. Math. Anal. Appl. 18 (1967), 135–144.

[3] Derndinger, R. & Nagel, R.: Der Generator stark stetiger Verbandshalbgruppen auf C(X) und dessen Spektrum, Math. Ann. 245 (1979), 159–177.

[4] Dubreil-Jacotin, M.: Sur le passage des équations de Navier-Stokes aux équations de Reynolds, C.R. Acad. Sci. Paris 244 (1957), 2887–2890.

[5] Fox, R.A.S. & Miller, J.B.: Averaging and Reynolds operators on Banach algebras III, J. Math. Anal. Appl. 24 (1968), 225–238.

[6] Gamlen, J.L.B. & Miller, J.B.: Averaging and Reynolds operators on Banach algebras II, J. Math. Anal. Appl. 23 (1968), 183–197.

[7] Goldstein, J.A.: “Semigroups of Linear Operators and Applications”, Oxford University Press, New York, 1985.

[8] Hille, E. & Phillips, R.S.: “Functional Analysis and Semigroups”, AMS Coll. Publ. 31, Providence (R.I.), 1957.

[9] Miller, J.B.: Möbius transforms of Reynolds operators, J. Reine Angew. Math. 218 (1965), 6–16.

[10] Miller, J.B.: A formula for the resolvent of a Reynolds operator, J. Aust. Math. Soc. 8 (1968), 447–456.

[11] Miller, J.B.: Averaging and Reynolds operators on Banach algebras I, J. Math. Anal. Appl. 14 (1966), 527–548.

[12] Molinaro, I.: Détermination d’une R-transformation de Reynolds, C.R. Acad. Sci. Paris 244 (1957), 2890–2893.

[13] Nagel, R. Ed.: “One-Parameter Semigroups of Positive Operators”, Springer, Berlin, 1985.

[14] Rota, G.-C.: Reynolds operators, Proc. Symp. Appl. Math. 16 (1962), 70–83.

[15] Sakai, S.: “Operator algebras in dynamical systems”, Cambridge University Press, Cambridge, 1991.

[16] Schaefer, H.H.: “Banach Lattices and Positive Operators”, Springer, Berlin, Heidelberg, New York, 1974.

[17] Scheffold, E.: Über Reynoldsoperatoren und Mittelwert bildende Operatoren auf halbeinfachen F-Banachverbandsalgebren, Math. Nachr. 162 (1993), 329–337.

[18] Stone, M.H.: A generalized Weierstrass Approximation Theorem. In: Studies in Modern Analysis, pp. 30–87. MAA Studies in Mathematics Vol. I, The Mathematical Association of America, Prentice-Hall, 1962.

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